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Natalka [10]
3 years ago
10

If the pressure of a gas is 2.000 × 104 kPa, what is its pressure in atm?

Chemistry
2 answers:
lisov135 [29]3 years ago
7 0

Answer:

197200atm

Explanation:

There are different units to express pressure of a gas, some of them are atmosphere, Bar, Kilopascal, Pascal, torr among other.

In this exercise we need to know the equivalence between atm and Kpa.

                                             if, 1kp=0,00986atm

then,

(2,000 x104kPa)*(0,00986atm)/1Kpa= 0,00986 atm

luda_lava [24]3 years ago
4 0

<u>Answer:</u>

<em>197.4 atm is the Answer.</em>

<em></em>

<u>Explanation:</u>

Pressure can be expressed in unit of torr, mm Hg, psi, atm, kPa etc

<em>1 atm = 760 torr = 760 mm Hg = 14.7 psi = 101.325 kPa </em>

So Here to convert atm to kPa the conversion factor is either \frac {1atm}{101.325kPa} or \frac {101.325kPa}{1atm}

We need the answer in atm so kPa should get cancel and it should be in the denominator

2.000 \times 10^4 kPa\times \frac {1atm}{101.325kPa}

=197.4 atm is the Answer.

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a bachelor's degree

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3 years ago
A 10 g sample of an unknown metal is cooled from 100 °C to 21.6 °C in a calorimeter of 100 g of water. The temperature of the wa
Marta_Voda [28]

Taking into account the definition of calorimetry, the specific heat of the unknown metal is 0.9 \frac{J}{gC} and the metal is aluminum.  

Calorimetry is the measurement and calculation of the amounts of heat exchanged by a body or a system.

Sensible heat is defined as the amount of heat that a body absorbs or releases without any changes in its physical state (phase change).

So, the equation that allows to calculate heat exchanges is:

Q = c× m× ΔT

where Q is the heat exchanged by a body of mass m, made up of a specific heat substance c and where ΔT is the temperature variation.

In this case, you know:

  • For unknown metal:
  1. Mass of metal = 10 g
  2. Initial temperature of metal= 100 °C
  3. Final temperature of metal= 21.6 ºC
  4. Specific heat of metal= unknown
  • For water:
  1. Mass of water = 100 g
  2. Initial temperature of water= 20 ºC
  3. Final temperature of water= 21.6 ºC
  4. Specific heat of water = 4.186 \frac{J}{gC}

Replacing in the expression to calculate heat exchanges:

For unknown metal: Qmetal= c_{unknown metal} × 10 g× (21.6 C - 100 C)

For water: Qwater= 4.186 \frac{J}{gC}× 100 g× (21.6 C - 20 C)

If two isolated bodies or systems exchange energy in the form of heat, the quantity received by one of them is equal to the quantity transferred by the other body. That is, the total energy exchanged remains constant, it is conserved.

Then, the heat that the gold gives up will be equal to the heat that the water receives. Therefore:

- Qmetal = + Qwater

- c_{unknown metal} × 10 g× (21.6 C - 100 C)=4.186 \frac{J}{gC}× 100 g× (21.6 C - 20 C)

Solving:

- c_{unknown metal} × 10 g× (- 78.4 C)= 669.76 J

c_{unknown metal} × 784 gC= 669.76 J

c_{unknown metal} = 669.76 J÷ 784 gC

c_{unknown metal} = 0.854 \frac{J}{gC} ≅ 0.9 \frac{J}{gC}

Finally, the specific heat of the unknown metal is 0.9 \frac{J}{gC} and the metal is aluminum.  

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- Calculate the Standard Enthalpy of the reaction below:
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Answer:

Explanation:

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NH_{3(g)} + HCl_{(g)} => NH_4Cl_{(s)}  .....     \Delta H = ? (1)\\2HCl_{(g)} => H_{2(g)} + Cl_{2(g)} .... \Delta H = +184.6 kJ (2)\\2H_{2(g)} + 1/2N_{2(g)} + 1/2Cl_{2(g)} => NH_4Cl_{(s)} ..... \Delta H = -314.2 kJ (3)\\ N_{2(g)} + 3H_{2(g)}  => 2NH_{3(g)} .... \Delta H  = +184.6kJ (4)

(can you double check that it is 184.6kJ for both equations 2 and 4 because it seems unlikely). We need to solve for equation 1 by addition and changing equations 2, 3 and 4. After possibly some trial and error, we can find that if we flip equations 4, multiply equation 3 by 2, add the equations together, and then finally divide by 2, we can get equation 1. We will get the answer of -314.2 kJ. However, I am again skeptical about the delta H values for equation 2 and 4 so double check that. This method might be super confusing and it is really hard to explain. So what I would suggest you to watch videos on Hess' law.

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3 years ago
How did greenhouse gasses get their name?
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Answer:

Greenhouse gasses got their name from a greenhouse

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Since greenhouses have plants inside in which the building lets sunlight in to the plants and greenhouse gasses trap heat. (sorry if my answer is confusing)

6 0
3 years ago
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Answer:

1 mole of Pbd - 239.3 g

1 mole  of H - 1.008 g

1 mole of Pb(NO3)2 - 331.2 g

So D

Explanation:

3 0
3 years ago
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